Questions on regulator circuit

Thread Starter

Zeebit

Joined Nov 8, 2013
72
I'm just starting to learn electronics as my hobby and as usual, I will start out by making my own power supply even if I don't know how the circuitry works.

What does R3 do? Say I change the current adjust pot to 150K to give a max current output of 3A, does R3 stay the same?

Correct link below.
 
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Metalmann

Joined Dec 8, 2012
703
I'm just starting to learn electronics as my hobby and as usual, I will start out by making my own power supply even if I don't know how the circuitry works.

http://www.quan-diy.com/projects/benchps/psschem.GIF

What does R3 do? Say I change the current adjust pot to 150K to give a max current output of 3A, does R3 stay the same?


Hilarious, except for the Rumsfeld (War Mongerer) part.
That guy, (and his buddies); needs tried for war crimes.
No if, ands, or buts.
 

Thread Starter

Zeebit

Joined Nov 8, 2013
72
What is your guess?

Also, why do you need 5A when you are just getting into electronics? eBay types of projects do you plan?
Like I said, I will make it into a 3A supply by changing the current adjust pot from 250k to 150k.

My guess is that R3 remains the same. I do not know why though.:p
 

Jony130

Joined Feb 17, 2009
5,487
R3 is current sensing resistor. The large the output current the large the voltage drop across this resistor. And when this voltage drop is large enough, the op amp comes to play and circuit act just like a current source (CC mode).
And this current is equal to
Isc = Vref* R2/(R3 + R5) + Vbe/R1 = 1.25V*R2/(R3+R5) + Vbe/R1 = 1.25V*150K/(330K * 0.2) + 0.7V/33 = 2.9A

In CV mode the output voltage is equal to Vout = 1.25V*(1+ R8/R6).
We have CV mode when I_load < Isc.
 
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Thread Starter

Zeebit

Joined Nov 8, 2013
72
Thanks for the explanation. I really have a lot of studying to do.

Also, based on the schematic, why is there a drop of 5V from input to output?
 

Dr.killjoy

Joined Apr 28, 2013
1,196
Thanks for the explanation. I really have a lot of studying to do.

Also, based on the schematic, why is there a drop of 5V from input to output?
Please don't build something or mod anything if you have no idea what your doing.... If you want to attempt something, please post the schematic for help because I would hate to hate see you get hurt,killed,or burn the house down because something happened....
 

Thread Starter

Zeebit

Joined Nov 8, 2013
72
What I'll be building is based entirely on the schematic on the OP. It's from the datasheet of National's LM317.

The only mod is changing R2 from 250K to 150K because my transformer can't handle 5A.
 

GopherT

Joined Nov 23, 2012
8,009
A linear voltage regulator will have a "drop out voltage". That is, how much extra input voltage is required for a given output. See datasheet.

This circuit can accept any input from about 6 to 35 volts . 35 is the upper limit of the regulator and op amp. If lower voltages are input, it limits the output range to something less.

Note that this circuit requires a -6 volt (negative 6 volt) supply for the op amp negative.
 

bountyhunter

Joined Sep 7, 2009
2,512
Bumping this thread.

Can the op amp be supplied with -5V instead of -6V on the datasheet?
Not sure, I doubt it. The output pin 6 of the LM301 has to pull about 4V below ground to be able to control the current in case the output is shorted to ground. Remember the adj pin of the 317 is always 1.25V negative of the OP, so if the OP is at ground, the ADJ pin will be at -1.25V. Then there is one diode drop for the 1N457 and another for the LED so that probably takes you down to about -4V ballpark. For the LM301 output to pull that low, it's negative supply has to be another 2V (at least) more negative.

I suspect that -6V number is there for that reason.
 
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